Evaluate Reverse Polish Notation
2014.2.25 23:42
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.
Some examples:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9 ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
Solution:
The Reverse Polish Notation is a postorder traversal of the syntax tree, which is constructed with operands and operators.
With a stack it would be easy to process the sequence and calculate the result.
Total time and space complexities are both O(n).
Accepted code:
1 // 1AC, simple training on stack operation. 2 #include <stack> 3 #include <vector> 4 using namespace std; 5 6 class Solution { 7 public: 8 int evalRPN(vector<string> &tokens) { 9 int i, n; 10 int op1, op2; 11 stack<int> nums; 12 bool is_op; 13 14 n = (int)tokens.size(); 15 for (i = 0; i < n; ++i) { 16 is_op = false; 17 if (tokens[i].length() == 1) { 18 switch(tokens[i][0]) { 19 case '+': 20 case '-': 21 case '*': 22 case '/': 23 is_op = true; 24 break; 25 } 26 } 27 28 if (is_op) { 29 if (nums.size() < 2) { 30 // not enough operands 31 return 0; 32 } 33 op2 = nums.top(); 34 nums.pop(); 35 op1 = nums.top(); 36 nums.pop(); 37 switch (tokens[i][0]) { 38 case '+': 39 nums.push(op1 + op2); 40 break; 41 case '-': 42 nums.push(op1 - op2); 43 break; 44 case '*': 45 nums.push(op1 * op2); 46 break; 47 case '/': 48 if (op2 == 0) { 49 // divide by 0 50 return 0; 51 } 52 nums.push(op1 / op2); 53 break; 54 } 55 } else { 56 if (sscanf(tokens[i].c_str(), "%d", &op1) != 1) { 57 // invalid integer format 58 return 0; 59 } 60 nums.push(op1); 61 } 62 } 63 int result = nums.top(); 64 nums.pop(); 65 66 return result; 67 } 68 };